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Let $k\geq 1$ a fixed integer and we put $u_n=\frac{1}{n+k}$ for all $n\in\mathbb{N}$ and we consider the sequence of matrices $$ M_0=(u_0) $$ and $$ M_1=\left(\begin{matrix}u_0 & u_1 \\ u_1 & u_2 \end{matrix}\right) $$ $$ M_2=\left(\begin{matrix}u_0 & u_1 & u_2\\ u_1 & u_2 &u_3 \\u_2 & u_3&u_4 \end{matrix}\right) $$ $$ M_3=\left(\begin{matrix}u_0 & u_1 & u_2&u_3\\ u_1 & u_2 &u_3 &u_4\\u_2 & u_3&u_4 &u_5 \\ u_3&u_4&u_5&u_6\end{matrix}\right) $$ and $$ M_n=(u_{i+j})_{i,j\leq n} $$ How to proof that for all $n\in\mathbb{N}$ $M_n$ is positive definite?

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  • $\begingroup$ Looks like a variation of hilbert matrix. $\endgroup$
    – user251257
    Aug 17, 2015 at 14:24
  • $\begingroup$ yeah for $k=1$ you can see Hilbert matrix $\endgroup$
    – Hamza
    Aug 17, 2015 at 14:28
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    $\begingroup$ Hint: $M_n$ is the Gramian matrix of $x^0, x^1, \dotsc, x^{n-1}$ to the inner product $\langle f, g \rangle = \int_0^1 f(x) g(x) x^{k-1} \; dx$. $\endgroup$
    – user251257
    Aug 17, 2015 at 14:31

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In $L^2([0,1])$ we consider the inner product : $$ \langle f, g \rangle=\int_0^1 f(t)g(t)t^{k-1} dt $$ so we can verify that $u_{n+m}=\langle x^n,x^m\rangle$ so for all $n\in \mathbb{N}$, $M_n$ is the gram matrix associated to this inner product.

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